What is the equation for the current in a LR circuit?
2 Answers
In an LR circuit, which consists of an inductor (L) and a resistor (R) connected in series with a voltage source, the behavior of the current over time can be described using a differential equation. Here’s a detailed breakdown of the equation governing the current:
### 1. **Basic Components:**
- **Inductor (L)**: This component opposes changes in current and stores energy in its magnetic field.
- **Resistor (R)**: This component opposes the flow of current and dissipates energy as heat.
### 2. **Applying Kirchhoff's Voltage Law:**
In a simple LR circuit connected to a constant voltage source \( V \), Kirchhoff's Voltage Law states that the sum of the potential differences (voltage) around a closed circuit must equal zero. Thus, we can express the voltage across the inductor and resistor as follows:
\[
V = V_R + V_L
\]
Where:
- \( V_R = iR \) (Ohm's law, where \( i \) is the current through the resistor)
- \( V_L = L \frac{di}{dt} \) (voltage across the inductor, where \( \frac{di}{dt} \) is the rate of change of current)
### 3. **Setting Up the Equation:**
Substituting these expressions into the voltage equation gives:
\[
V = iR + L \frac{di}{dt}
\]
Rearranging this yields:
\[
L \frac{di}{dt} + iR = V
\]
### 4. **Solving the Differential Equation:**
This is a first-order linear ordinary differential equation. To solve it, we can rearrange it as:
\[
\frac{di}{dt} + \frac{R}{L} i = \frac{V}{L}
\]
This is a standard form that can be solved using an integrating factor or by recognizing it as a linear differential equation.
### 5. **Current Response:**
The solution to this equation shows how the current changes over time as the circuit reaches a steady state. The general solution for the current \( i(t) \) in the circuit can be expressed as:
\[
i(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right)
\]
### 6. **Understanding the Equation:**
- **Steady-State Current**: As \( t \) approaches infinity, \( e^{-\frac{R}{L}t} \) approaches zero, and the current approaches its maximum value, \( \frac{V}{R} \). This represents the steady-state current when the circuit is fully energized.
- **Time Constant (\( \tau \))**: The term \( \frac{L}{R} \) is known as the time constant \( \tau \). It indicates how quickly the current rises to its steady state. A larger \( \tau \) means a slower rise in current.
### 7. **Conclusion:**
In summary, the current in an LR circuit increases exponentially from zero to a maximum value as it responds to a step voltage input. The equation for current is:
\[
i(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right)
\]
This equation captures the transient behavior of the circuit and shows how the inductor and resistor work together to shape the current response over time.
### 1. **Basic Components:**
- **Inductor (L)**: This component opposes changes in current and stores energy in its magnetic field.
- **Resistor (R)**: This component opposes the flow of current and dissipates energy as heat.
### 2. **Applying Kirchhoff's Voltage Law:**
In a simple LR circuit connected to a constant voltage source \( V \), Kirchhoff's Voltage Law states that the sum of the potential differences (voltage) around a closed circuit must equal zero. Thus, we can express the voltage across the inductor and resistor as follows:
\[
V = V_R + V_L
\]
Where:
- \( V_R = iR \) (Ohm's law, where \( i \) is the current through the resistor)
- \( V_L = L \frac{di}{dt} \) (voltage across the inductor, where \( \frac{di}{dt} \) is the rate of change of current)
### 3. **Setting Up the Equation:**
Substituting these expressions into the voltage equation gives:
\[
V = iR + L \frac{di}{dt}
\]
Rearranging this yields:
\[
L \frac{di}{dt} + iR = V
\]
### 4. **Solving the Differential Equation:**
This is a first-order linear ordinary differential equation. To solve it, we can rearrange it as:
\[
\frac{di}{dt} + \frac{R}{L} i = \frac{V}{L}
\]
This is a standard form that can be solved using an integrating factor or by recognizing it as a linear differential equation.
### 5. **Current Response:**
The solution to this equation shows how the current changes over time as the circuit reaches a steady state. The general solution for the current \( i(t) \) in the circuit can be expressed as:
\[
i(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right)
\]
### 6. **Understanding the Equation:**
- **Steady-State Current**: As \( t \) approaches infinity, \( e^{-\frac{R}{L}t} \) approaches zero, and the current approaches its maximum value, \( \frac{V}{R} \). This represents the steady-state current when the circuit is fully energized.
- **Time Constant (\( \tau \))**: The term \( \frac{L}{R} \) is known as the time constant \( \tau \). It indicates how quickly the current rises to its steady state. A larger \( \tau \) means a slower rise in current.
### 7. **Conclusion:**
In summary, the current in an LR circuit increases exponentially from zero to a maximum value as it responds to a step voltage input. The equation for current is:
\[
i(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right)
\]
This equation captures the transient behavior of the circuit and shows how the inductor and resistor work together to shape the current response over time.
In an LR circuit, which consists of an inductor \(L\) and a resistor \(R\) connected in series, the current \(I(t)\) as a function of time \(t\) when a constant voltage \(V\) is applied can be described by the following differential equation:
\[ V = L \frac{dI(t)}{dt} + IR \]
To find the current \(I(t)\), solve this differential equation. The solution depends on whether the circuit is starting from rest or has a different initial condition.
### 1. **For a Circuit Starting from Rest**
If the circuit starts with no initial current (i.e., \(I(0) = 0\)), the solution to the differential equation is:
\[ I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
- \(V\) is the constant applied voltage.
- \(R\) is the resistance.
- \(L\) is the inductance.
- \(t\) is time.
### 2. **For a Circuit with Initial Current \(I_0\)**
If the circuit has an initial current \(I_0\) at \(t = 0\), the solution is:
\[ I(t) = \frac{V - I_0 R}{R} \left(1 - e^{-\frac{R}{L}t}\right) + I_0 \]
- \(I_0\) is the initial current in the circuit.
### Explanation
- **Exponential Term:** The term \(e^{-\frac{R}{L}t}\) represents the transient response of the circuit. It shows how the current builds up over time as the inductor's initial opposition to current changes.
- **Steady-State Current:** As \(t \to \infty\), the exponential term \(e^{-\frac{R}{L}t}\) approaches zero, and the current \(I(t)\) approaches the steady-state value \( \frac{V}{R} \).
This describes how the current increases from zero to its final steady-state value in an LR circuit with a step input voltage.
\[ V = L \frac{dI(t)}{dt} + IR \]
To find the current \(I(t)\), solve this differential equation. The solution depends on whether the circuit is starting from rest or has a different initial condition.
### 1. **For a Circuit Starting from Rest**
If the circuit starts with no initial current (i.e., \(I(0) = 0\)), the solution to the differential equation is:
\[ I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
- \(V\) is the constant applied voltage.
- \(R\) is the resistance.
- \(L\) is the inductance.
- \(t\) is time.
### 2. **For a Circuit with Initial Current \(I_0\)**
If the circuit has an initial current \(I_0\) at \(t = 0\), the solution is:
\[ I(t) = \frac{V - I_0 R}{R} \left(1 - e^{-\frac{R}{L}t}\right) + I_0 \]
- \(I_0\) is the initial current in the circuit.
### Explanation
- **Exponential Term:** The term \(e^{-\frac{R}{L}t}\) represents the transient response of the circuit. It shows how the current builds up over time as the inductor's initial opposition to current changes.
- **Steady-State Current:** As \(t \to \infty\), the exponential term \(e^{-\frac{R}{L}t}\) approaches zero, and the current \(I(t)\) approaches the steady-state value \( \frac{V}{R} \).
This describes how the current increases from zero to its final steady-state value in an LR circuit with a step input voltage.